Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 15 / c / Solution

Solving for $y$ gives $y=\pm (b/a)\sqrt{x^2-a^2}$ with $x\ge a$. For large $x$, the square root is $x-a^2/(2x)+O(x^{-3})$, so the two asymptotes areThe curve lies between these lines on the right side and has its nearest point, with vertical tangent, at $(a,0)$. As $t$ increases it travels from the lower unbounded end through that vertex to the upper unbounded end. The left hyperbola branch is not part of the given curve.